Homework Help: Integral in the form 1/sqrt

1. Jun 21, 2007

sokratesla

integral in the form dx/sqrt(x^2 + a^2)

1. The problem statement, all variables and given/known data
# Hi, this is my first post!
# I have to solve an integral in my electrodynamics study for evaluation of an electric potential. They are in the form $$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int \rho(\mathbf{r'})/\abs(\mathbf{r}-\mathbf{r'}) d\mathbf{r'}$$ In order to find the potential on the top of a conical surface I have to solve an integral in the following form:

2. Relevant equations
$$\int \frac{dx}{\sqrt{x^2+a^2}}$$

3. The attempt at a solution
I solved another problem of the kind
$$\int \frac{x dx}{\sqrt{x^2+a^2}}$$
but stucked with this one.
Can you give a hint, or help in order to solve this integral.

http://integrals.wolfram.com/index.jsp
Mathematica gives $$\log(x+\sqrt{x^2+a^2})$$

Last edited: Jun 21, 2007
2. Jun 21, 2007

malawi_glenn

Okay, the attempt at a solution?

Do you own a handbook on mathematical identites, have you search the internet and/or your books on one-dimensional calculus?

3. Jun 21, 2007

bel

Have you tried any trigonometric or hyperbolic function substutions? Might work out to be arcosh or arsinh of some not so pretty thing, my guess.

Let's see, int(1/sqrt(x^2+a^2))dx = arsinh (x/a) , a>0

Last edited: Jun 21, 2007
4. Jun 21, 2007

malawi_glenn

yes that is correct; $$\ln(x+\sqrt{x^2+a^2})$$
it is.

5. Jun 21, 2007

sokratesla

# Thanks for the reply!
# Here is my attempt. I found this substitution in my Calculus Book and thought it may be useful.
$$x\equiv a\tan(\theta)$$
so the denominator becomes:
$$a\sqrt{\tan^2(\theta)+1}$$ and integral becomes $$\frac{\cos{\theta}dx}{a}$$
$$dx=a\sec^2(\theta)$$
and at last the integral after substitution is:
$$\int \frac{d\theta}{\cos{\theta}}$$

Last edited: Jun 21, 2007
6. Jun 21, 2007

malawi_glenn

Okay, now I am confused what you want to calculate.

$$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int ..$$

or

$$\int \frac{dx}{\sqrt{x^2+a^2}}$$

??

What is the form of the charge distribution?

7. Jun 21, 2007

sokratesla

# Interesting, Maxima gives this as an answer. Mathematica and Maxima seems to give different answers but, then, may be this expressions are equal.

8. Jun 21, 2007

sokratesla

# Second one. First one is the general formula for vector potentials. But I reduced my problem to the second integral.

9. Jun 21, 2007

malawi_glenn

Okay, then:

$$\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2})$$

10. Jun 21, 2007

VietDao29

Yes, whenever you see something like x2 + a2, the first thing you'd like to try would be the substitution: x = atan(t).

After arriving at:
$$\int \frac{d (\theta)}{\cos \theta}$$, we notice that the power of the cosine function is 1, ie, odd, so, we'll use the u-substitution: $$u = \sin \theta$$, like this:
$$\int \frac{d (\theta)}{\cos \theta} = \int \frac{\cos \theta d (\theta)}{\cos ^ 2 \theta} = \int \frac{\cos \theta d (\theta)}{1 - \sin ^ 2 \theta} = \int \frac{du}{1 - u ^ 2}$$.
Hopefully, you can go from here, right?

----------------------------

Or, a different way to tackle this integral is to use Euler Substitution.
Let $$t = \sqrt{x ^ 2 + a ^ 2} + x$$
Take the differential of both sides yields:
$$dt = \frac{x}{\sqrt{x ^ 2 + a ^ 2}} dx + dx = \left(\frac{x}{\sqrt{x ^ 2 + a ^ 2}} + 1 \right) dx = \frac{x + \sqrt{x ^ 2 + a ^ 2}}{\sqrt{x ^ 2 + a ^ 2}} = \frac{t}{\sqrt{x ^ 2 + a ^ 2}} dx$$

Isolate x's, and t's to one side, we have:
$$\Rightarrow \frac{dt}{t} = \frac{dx}{\sqrt{x ^ 2 + a ^ 2}}$$
Now, integrate both sides gives:
$$\int \frac{dx}{\sqrt{x ^ 2 + a ^ 2}} = \int \frac{dt}{t} = \ln|t| + C = \ln|x + \sqrt{x ^ 2 + a ^ 2}| + C$$

----------------------------

Or, to make it more convenient, you can try to learn this integral by heart, it's pretty common.

$$\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2}) + C$$

Last edited: Jun 21, 2007
11. Jun 21, 2007

George Jones

Staff Emeritus
y = arcsinh(x/a)

sinhy = x/a

(e^y - e^-y)/2 = x/a

Multiply by e^y and rearrange.

e^(2y) - 2x/a e^y - 1 = 0

Use the formula for the roots of a quadratic and simplify.

e^y = (x +/- sqrt(x^2 + a^2))/a

y = ln(x + sqrt(x^2 + a^2) - lna

So, up to a constant of integration, the two answers are the same.

12. Jun 21, 2007

sokratesla

# Yes, it is the answers. But everywhere what I found is the answer. But I want to learn the intermediate steps too. Otherway I'll feel like I'm cheating in the exam if I memorize the table without learning its proof.
http://www.sosmath.com/tables/integral/integ11/integ11.html
# Here there are two answers.

# I tried the substitution x=a*tan(u) and changed the integral to du / cos(u) but stopped there.

13. Jun 21, 2007

malawi_glenn

hehe yeah, but is it an exam on integral calculus or electrodynamics?..
Are you not allowed to use collection of formulas and so on?

14. Jun 21, 2007

bel

let x = a sinh (psi)
dx = a cosh (psi) d(psi)

then int(1/sqrt(x^2+a^2))dx = int (acosh(psi)/ sqrt((a^2)(sinh(psi))^2 + a^2)) d(psi)

= int (a cosh (psi)/ a cosh (psi) ) d(psi)
= int (1) d(psi)
= psi+C

but, x = a sinh (psi)
sihn (psi)= (x/a)
(psi) = arsinh (x/a)

so, int(1/sqrt(x^2+a^2))dx = psi+C = arsinh (x/a) +C

my guess is that ln(x+ sqrt (x^2+a^2)) and this differs by a constant.

sorry I don't know how to type maths symbols, or this would have looked nicer.

15. Jun 21, 2007

malawi_glenn

16. Jun 21, 2007

bel

thanks, malawi_glenn. =)

but before I learn all that, the ln(u) thing probably came from the fact that sinh(u)+cosh(u)=e^(u).

17. Jun 21, 2007

sokratesla

# Thanks for these important advises. du/(1-u^2) seems much simpler. I think I can solve it looking my Calculus book.

# And many thanks to everybody! Have a nice day!

18. Jun 21, 2007

sokratesla

# It's and electrodynamics exam. In general, the exams are openbook or instructors give some important formulas. But they are all physics books and formulas.
# And, I think, it's a shame for last year physics students, like me, not able to solve basic integrals.:shy:
# Thanks for your help.

19. Jun 21, 2007

bob1182006

Actually this is true for
$$\int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln(x+\sqrt{x^2 \pm a^2})$$

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